THe figure above shows a small mass connected to a string, which is attached to a vertical post. If the mass is released when the string is horizontal as shown, the magnitude of the total acceleration of the mass as a function of the angle is

Mechanics}Boundary Condition

Getting low on time, one should begin scoring points based more of testing strategy than sound rigorous physics. At the initial release point, the acceleration is due to gravity and the tension is 0 (no centripetal acceleration). The only choice that gives is choice (E).

Alternate Solutions

his dudeness2010-09-05 15:55:11

Yosun's method is clearly by far the best for this particular problem. However, if anyone is curious about how the relation in (E) is actually derived, here you go:

The acceleration can be split into radial and transverse components. Since gravity is the only force contributing to transverse acceleration, We have . To calculate , we note that . From conservation of energy, we have , so we get .

A real way to do the problem is as follows. Kinetic energy is , where is the length of the string; potential is . Setting the Lagrangian to , Lagrange's equation is . In our case and we find , which then yields . But this is only the acceleration along the direction of motion; there is also a perpendicular centripetal acceleration given by . To find we can invoke conservation of energy: , so , so the centripetal acceleration is .

Taking the norm of the the two acceleration components, we get , answer (E).

A common mistake would be to omit the calculation of the centripetal acceleration, which would yield an incorrect answer of , which ETS kindly does not give as an option.

Yosun's method is clearly by far the best for this particular problem. However, if anyone is curious about how the relation in (E) is actually derived, here you go:

The acceleration can be split into radial and transverse components. Since gravity is the only force contributing to transverse acceleration, We have . To calculate , we note that . From conservation of energy, we have , so we get .

All together now: . A simple trig identity gets us the (E).

physicsres.com2014-11-15 12:55:37

good answer just a typo alert.

unurbat2016-06-28 23:24:16

Where did the \"r\" disappear when you try to find radial acceleration?

Oh Jesus, people, get all this nonsense away. Calculations on the GRE? That's suicide!

Here's how you think of it:

When is , the mass wants to go straight down, so a = g.

That's it. Only E survives that.

But, if you had a different choice of answers, you can also think that acceleration is a maximum at , because it also has centripetal acceleration, which adds to the gravitational force at that point. This obviously corresponds to sin, so there you go.

ajkp25572009-11-03 13:57:32

"Calculations on the GRE? That's suicide!"

Best. Advice. Ever. The PGRE is entirely about test taking strategy. Yes, once you are in grad school it will be very important to know how to correctly work every problem, but spending four minutes on a problem on a test that only gives you 1.7 minutes per problem isn't a smart approach. You get the same credit for the answer, regardless of how you got it.

I don\'t understand why the acceleration is 2g at the bottom. The total acceleration is the combination of g, which points down and the centripetal acceleration which points up, so it must be less than g total, right?

A real way to do the problem is as follows. Kinetic energy is , where is the length of the string; potential is . Setting the Lagrangian to , Lagrange's equation is . In our case and we find , which then yields . But this is only the acceleration along the direction of motion; there is also a perpendicular centripetal acceleration given by . To find we can invoke conservation of energy: , so , so the centripetal acceleration is .

Taking the norm of the the two acceleration components, we get , answer (E).

A common mistake would be to omit the calculation of the centripetal acceleration, which would yield an incorrect answer of , which ETS kindly does not give as an option.

kroner2009-10-06 22:29:16

You don't need to do all that work to find the tangential acceleration. It's just the tangential component of gravity, which is .